data structures chapter 10: efficient binary search trees 10-1
TRANSCRIPT
Data StructuresChapter 10: Efficient Binary Search Trees
10-1
Two Binary Search Trees (BST)
• For each identifier with equal searching probability , average # of comparisons for a successful search:Left: (1+2+2+3+4)/5 = 2.4
Right: (1+2+2+3+3)/5 = 2.2
• prob(5, 10, 15, 20, 25) = (0.3, 0.3, 0.05, 0.05, 0.3)Left :0.3(2+1+2)+0.05(4+3) = 1.85
Right: 0.3(2+1+3)+0.05(3+2) = 2.05
10
25
20
15
5
10
20
15 25
52 2 2 2
3 3 3
1 1
4
10-2
Extended Binary Trees• Add external nodes to the original
binary search tree– Take external nodes as failure nodes
.
• External / internal path length – Internal path length, I, is:
I = 0 + 1 + 1 + 2 + 3 = 7– External path length, E, is :
E = 2 + 2 + 4 + 4 + 3 + 2 = 17• E = I + 2n, where n is # of internal
nodes.
10
25
20
15
52
1
3
4
0
1
2
4
3
2 2
10-3
Search Cost in a BST• In the binary search tree( BST):
– Identifiers a1, a2, …, an with a1 < a2 < … < an
– pi :probability of successful search for ai
– qi : probability of unsuccessful search ai < x < ai+1
• Total cost
• An optimal binary search tree for a1, …, an is the one that minimizes the total cost.
n
ii
n
ii qp
01
1
n
ii
n
iii ilevelqalevelp
01
)1) node failure(()(
10
25
20
15
53
2
4
5
1
2
3
5
4
3 3
10-4
Algorithm for Constructing Optimal BST (1)• Solved by dynamic programming.
• Tij : an optimal binary search tree for ai+1, …, aj, i < j.
– Tii is an empty tree for 0 i n.
• cij : cost of Tij, where cii=0.
• rij : root of Tij
• wij : weight of Tij ,
• T0n is an optimal binary search for a1, …, an. cost: c0n weight: w0n root: r0n
j
ikkkiij pqqw
1
)(
10-5
• Suppose ak is the root of Tij (rij = k).• T has two subtrees L and R.
– L: left subtree with ai+1, …, ak1
– R: right subtree with ak+1, …, aj
cij = pk + cost(L) + cost(R) + weight(L) + weight(R) = pk + ci,k1 + ckj + wi,k1 + wkj
= wij + ci,k1 + ckj (wij = pk + wi,k1 + wkj)= wij +
• Time complexity: O(n3)
Algorithm for Constructing Optimal BST (2)
}{min 1, ljlijli
cc ak
L Rai+1, …, ak1 ak+1, …, aj
10-6
Example for Constructing Optimal BST (1)• n = 4, (a1, a2, a3, a4) = (10, 15, 20, 25).
16(p1, p2, p3, p4) = (3, 3, 1, 1) 16(q0, q1, q2, q3, q4) = (2, 3, 1, 1, 1).
• Initially wii = qi , cii = 0, and rii = 0, 0 i 4w01 = p1 + w00 + w11 = p1 + q0 +q1 = 8c01 = w01 + min{c00 +c11} = 8, r01 = 1
w12 = p2 + w11 + w22 = p2 + q1 + q2 = 7c12 = w12 + min{c11 +c22} = 7, r12 = 2
w14 = 11c14 = w14 + min{c11 +c24 , c12 +c34 , c13 +c44} = 11+ c11 +c24 =19, r14 = 2
w04 = 16c04 = w04 + min{c00 +c14 , c01 +c24 , c02 +c34 , c03 +c44} = 16+ c01 +c24 =32, r04 = 2
10-7
Example for Constructing Optimal BST (2)
• wii = qi
• wij = pk + wi,k1 + wkj
• cij = wij +• cii = 0• rii = 0• rij = l
Computation is carried out row-wise from row 0 to row 4.
}{min 1, ljlijli
cc
The optimal binary search tree
1
2
3
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(a1, a2, a3, a4) = (10, 15, 20, 25)
(p1, p2, p3, p4) = (3, 3, 1, 1) (q0, q1, q2, q3, q4) = (2, 3, 1, 1, 1)
15
20
25
10
w00=2c00=0r00=0
w11=3c11=0r11=0
w22=1c22=0r22=0
w00=2c00=0r00=0
w33=1c33=0r33=0
w44=1c44=0r44=0
w01=8c01=8r01=1
w12=7c12=7r12=2
w23=3c23=3r23=3
w34=3c34=3r34=4
w02=12c02=19r02=1
w13=9c13=12r13=2
w24=5c24=8r24=3
w03=14c03=25r03=2
w14=11c14=19r14=2
w04=16c04=32r04=2
0 1 2 3 4
4
0
1
2
3
10-8
AVL Trees• Height balanced binary search trees.• Proposed by G. Adelson-Velsky and E. M. Landis• Balance factor of each node v
– BF(v) =hL hR = 1, 0, or 1
– hL: height of left subtree
– hR: height of right subtree
• We can insert an element into the tree, or delete an element from it, in O(log n) time.
• At most one single rotation or double rotation is needed when an insertion is performed.
10-9
JAN
APR
AUG
DEC
SEPT
OCT
NOV
FEB MAR
MAYJUNE
JULY
Not an AVL tree:
An AVL tree:
JAN
DEC MAR
AUG FEB JULY NOV
-1
+1
0 -1OCT
-1
MAY
0
JUNE0APR
-1-1
0SEPT
0+1
10-10
Four Kinds of Rotations in an AVL Tree
• 4 rotations for rebalancing: LL, RR, LR, and RL
• These rotations are characterized by the nearest ancestor A of the inserted node Y whose balance factor becomes 2.– LL: insert new node Y in the left subtree of the left subtree of A.– RR: insert Y in the right subtree of the right subtree of A– LR: insert Y in the right subtree of the left subtree of A– RL: insert Y in the left subtree of the right subtree of A
• LL and RR are symmetric, called single rotations.
• LR and RL are symmetric, called double rotations.
10-11
LL Rebalancing Rotation+1A
0B
BL
hBR
h
AR
hh+2
+2A
+1B
BL
h+1BR
h
AR
h
0B
BL
h+1
0A
BR
hAR
h
h+2
LL
height of BL increases to h+1, right rotation
(e) Insert APR
MAY+2
NOV
0+2
MAR+1
AUG0
APR
LL MAY+1
NOV
00
AUG0
APR MAR
0
10-12
RR Rebalancing Rotation-1A
0B
BR
hBL
h
AL
h
h+2
-2A
-1B
BR
h+1BL
h
AL
h
0B
BR
h+1
0A
BL
hAL
h
h+2
RR
APR
AUG FEB
JAN
DEC MAR
JULY MAY
NOV
OCT
JUNE
-1
+1 -1
-1
-1
-2
0
0
0+1
0
(k) Insert OCT
APR
AUG FEB
JAN
DEC MAR
JULY NOV
MAY OCTJUNE
-1
+1 0
-1
0
0
00
0+1
0
RR
height of BR increases to h+1, left rotation
10-13
LR Rebalancing Rotation
CL
h
+1A
0B
0C
BL
hCL
h-1CR
h-1
AR
h
h+2
+2A
-1B
+1C
BL
hCL
hCR
h-1
AR
h
LR 0C
0B
-1A
BL
h
CR
h-1AR
h
h+2
• LR needs double rotations:1. Perform left rotation on the tree rooted at B.
2. Perform right rotation on the tree rooted at A.
10-14
Example of LR
(f) Insert JAN
MAY
+2
NOV
0-1
AUG0
APR MAR
+1
0
JAN
MAR
0
MAY
-10
AUG0
APR JAN
0
NOV
0
LR
10-15
Complexity Comparison of Various Structures
Operation Sequential List (Sorted Array)
Linked List
AVL Tree
Search for x O(log n) O(n) O(log n)Search for kth item O(1) O(k) O(log n)
Delete x O(n) O(1)1 O(log n)
Delete kth item O(n k) O(k) O(log n)
Insert x O(n) O(1)2 O(log n)
Output in order O(n) O(n) O(n)
1Doubly linked list and position of x known.2Position for insertion known
10-16
Red-Black Trees• A red-black tree is an extended binary search tree.• Each node/pointer (edge) is colored by red or black.
– Colored nodes definition– Colored edges definition
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4 9
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1
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12107
Extended binary search tree
External nodes
Internal nodes
10-17
Colored Node Definition
• Colored node definition– RB1: The root and all external nodes are black.
– RB2: No root-to-external-node path has two consecutive red nodes.
– RB3: All root-to-external-node paths have the same number of black nodes.
10-18
65
50 80
10 60 70
5 62
Colored Pointer (Edge) Definition
• Colored pointer (edge) definition– RB1’: Pointer to an external node is black.– RB2’: No root-to-external-node path has two consecutive red
pointers (edges).– RB3’: All root-to-external-node paths have the same number of
black pointers.
10-19
65
50 80
10 60 70
5 62
Length and Rank in a Red-Black Tree• Let the length of a root-to-external-node path be the
number of pointers (edges) on the path.• Let the rank of a node be the number of black pointers
(edges) (equivalently the number of black nodes minus 1) on any path from the node to any external node in its subtree.
• Lemma: P, Q: two root-to-external-node paths
length(P) 2 * length (Q)• Proof: Let the rank of the root be r.
– From RB2’, each red pointer is followed by a black pointer.
– Therefore, each root-to-external-node path has between r and 2r pointers.
10-20
Properties of a Red-Black Tree
• Lemma: Let h be the height of a red-black tree (excluding the external nodes), let n be the number of internal nodes, and let r be the rank of the root.
(a) h 2r(b) n 2r1(c) h 2 log2(n+1)
• Proof: (a) is correct by previous lemma.
From (b), we have r log2(n+1). This inequality together with (a) yields (c).
• Height of a red-black tree 2 log2(n+1), searching, insertion, and deletion needs O(log n) time.
10-21
Inserting into a Red-Black Tree• A new element is first inserted as
the ordinary binary search tree.
• Assign the new node to red.
• The new node may or may not violate RB2 (imbalance).– One root-to-external-node path may
have two consecutive red nodes.– It can be handled by changing
colors or a rotation.
10-22
a b
c
d
gu
pu
u
Two Consecutive Red Nodes• u: new node, red • pu: parent of u, red • gu :grandparent of u, black • LLb, LLr
– Left child, then left child– LLb: the other child of gu, d, is black.– LLr: the other child of gu, d, is red.
• LRb, LRr: – Left child, then right child, black or red
• RRb, RRr: – Right child, then right child, black or red
• RLb, RLr: – Right child, then left child, black or red
a b
c
d
gu
pu
u
10-23
• Change color
• Move u, pu, and gu up two levels.– gu becomes new u
• Continue rebalancing if necessary.– If RB2 is satisfied, stop propagation.
– If gu is the root, force gu to be black (The number of black nodes on all root-to-external-node paths increases by 1.)
– Otherwise, continue color change or rotation.
Color Change of LLr, LRr, RRr, RLr
a b
cd
gupu
uLLr
a b
cd
gupu
u
10-24
red black
Rotation and Color Change of LLb, LRb, RRb, RLb
• Same as the rotation schemes taken for an AVL tree.• LLb rotation:
LL rotation of AVL tree
• LRb rotation: LR rotation of AVL tree
• RRb is symmetric to LLb
• RLb is symmetric to LRb.
a b c d
y
x z
c
d
gu
pu
u xx
z
y LLbLLb
u
a b
y
a b
z
c d
x
y
b c
a
d
gu
pu
u
x
z
LRbLRb
u
10-25
Inserting 50, 10, 80, 90, 70, 60, 65, 6250
Insert 50Insert 10
50
Insert 80
10 10 80
50
Insert 90
10 80
50
90
RRrRRru
pu
gu
d
10
50
90
u
pu
gu
d
80 root 10
50
90
u
pu
gu
d
80
This violates RB1 10-26
10
50
90
u
pu
gu
d
Insert 70
70
80
10
50
90
Insert 60
70
80
60
u
pu
10
50
90
80
60
LLr
70
Inserting 50, 10, 80, 90, 70, 60, 65, 62
10
50
90
80
10-27
u
pu
gu
d
10
50
90
80
60
70
Insert 65
65
LRbLRb
10
50
90
80
60
65
70
pu
u
gu
Inserting 50, 10, 80, 90, 70, 60, 65, 62
10-28
10
50
90
80
60
65
70
Insert 62
62
u
pu
gu
d LRrLRr
10
50
90
80
65
70
Insert 62 …
62
u
pu
gu
d
60
Inserting 50, 10, 80, 90, 70, 60, 65, 62
10-29
10
50
90
80
65
70
62
u
pu
gu
d
60
LRbLRb
10 90
65
70
62
u
60
8050
Further Process for Inserting 62
10-30
Splay Trees 斜張樹 /伸展樹• A splay tree is a binary search tree.• In an AVL tree, we have to store the balance
factor. In a red-black tree, we have store the red/black color.
• In a splay tree, there is no balanced information.• The operation for searching, insertion or deletion
needs O(log n) amortized time. (worst case O(n).)• Two varieties
– Bottom-up splay tree– Top-down splay tree
10-31
Bottom-Up Splay Trees• Searching, insertion and deletion are performed
as in an unbalanced binary search tree, then followed by a splay operation (a sequence of rotations).
• The start node x for the splay:– The searched, inserted node– The parent of the deleted node.
• After the splay operation completes, the splay node x becomes the tree root.
10-32
The Splay Operation• q: the start node for the splay• p: parent node of q• gp: grandparent of q • (1) If q=0 or q=root, then stops.• (2) If there is no gp, then perform a rotation.
10-33
xa
b c
p
q
x
a b
cp
q
a, b, and c are substrees
Rotations in the Splay Operation• (3) If q has a parent p and a grandparent gp, then
a rotation is performed: – LL: left child, left child
– RR: right child, right child
– LR; left child, right child
– RL: right child, left child
• Move up 2 levels at a time.• The splay is repeated at the new location of q,
until q becomes the root.• LL and RR are symmetric. LR and RL are
symmetric. 10-34
RR and RL Rotations
x
a
b
c d
gp x
d
c
ba
p
q
p
q
gp
x
b c
d
ax
a b c d
gp
p
q gp p
q
10-35
• RR rotation– Keep inorder
sequence unchanged
• RL rotation– Keep inorder
sequence unchanged
Example for the Splay Operation (1)
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8
2
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fe
d
c
b
a
g
h
i
j
(a) Initial search tree, RR
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5
b
a
g
h
i
j
4
3
dc
f
e
(b) After RR rotation, LL 10-36
Example for the Splay Operation (2)
1
9
8
2
5
a
i
j
4
3
dc
e
6
7
g h
f
b
(c) After LL rotation, LR (d) After LR rotation, RL
1
9
5
2
a
j
4
3
b
dc
e
8
6
7
i
g h
f
10-37
Example for the Splay Operation (3)
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2
4
e
b
5
3
dc
a
9
8
6
f
i
7
g h
j
(e) After RL rotation
10-38
Top-Down Splay Trees (1)• The splay node x (same as bottom-up splay
tree):– The searched, inserted node– The parent of the deleted node.
• Following the path from the root to the splay node, partition the binary search tree into 3 components:– small binary search tree (smaller than x)– big binary search tree (bigger than x)– the splay node x
10-39
Top-Down Splay Trees (2)• Move down 2 levels at a time, except
(possibly) that one level is moved down when the splay node is reached.
• A rotation is done whenever an LL or RR move is performed.
• When the splay node is reached, the small tree and the big tree are combined into a new binary search tree rooted at the splay node x.
10-40
An Example for Top-Down Splay Tree (1/7)
bigsmall1
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3
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5
fe
d
c
b
a
g
h
i
j
Initial search tree, RL (right subtree, then left subtree)
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a
9
j
s b
x
10-41
bigsmall
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2
7
6
3
4
5
fe
d
c
b
g
h
i
After RL transformation, LR (left, then right)
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a
9
j
s bx
2
b
8
i
An Example for Top-Down Splay Tree (2/7)
10-42
bigsmall
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6
3
4
5
fe
d
c
g
h
1
2
b
a
9
8
i
j
s b
x
After LR transformation, LL
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g h
An Example for Top-Down Splay Tree (3/7)
10-43
bigsmall
3
4
5
fe
d
c
1
2
b
a
9
8
6 i
7
g h
j
b
x
After LL transformation (a rotation), RR
s
4
3
dc
An Example for Top-Down Splay Tree (4/7)
10-44
bigsmall
5
fe
1
2
4b
3
dc
a
9
8
6
e f
i
7
g h
j
s b
x
After RR transformation (a rotation), splay node is reached
An Example for Top-Down Splay Tree (5/7)
10-45
bigsmall
5
1
2
4
e
b
3
dc
a
9
8
6
f
i
7
g h
j
s b
Splay node
x
5
An Example for Top-Down Splay Tree (6/7)
10-46
1
2
4
e
b
5
3
dc
a
9
8
6
f
i
7
g h
j
Final new search tree
An Example for Top-Down Splay Tree (7/7)
• Bottom-Up v.s. Top-Down– Top-down splay trees are faster than bottom-up splay trees
by experiments. 10-47